Matrices whose Linear Combinations are All Singular

Solution 1:

This is not a complete answer, just some sufficient and some necessary conditions.

Let's look at the opposite condition. Fix $u,v\in\mathrm{End}_k(V)$ two singular (non proportional) endomorphisms of the finite dimensional $k$-vector space $V$: what conditions are necessary or sufficient for there to be a scalar $\lambda$ such that $u+\lambda v\in\mathrm{GL}(V)$?

The following conditions are necessary:

1. $\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)$;
2. $\mathrm{Im}(u)+\mathrm{Im}(u)=V$;
3. $v\big(\mathrm{Ker}(u)\big)\oplus u\big(\mathrm{Ker}(v)\big)$.

The first is @chaochuang's answer, and the second is @Marc van Leeuwen' comment. To see the necessity of the third condition, write down matrices in a basis adapted to $V=\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)\oplus S$. The following conditions are sufficient (at least when $k$ is big enough, for instance if $\mathbb Q\subset k$):

1. The induced morphism $\tilde{v}:\mathrm{Ker}(u)\rightarrow V\rightarrow V/\mathrm{Im}(u)$ is an isomorphism i.e. $v(\mathrm{Ker}(u))\oplus\mathrm{Im}(u)$;
2. The induced morphism $\tilde{u}:\mathrm{Ker}(v)\rightarrow V\rightarrow V/\mathrm{Im}(v)$ is an isomorphism i.e. $u(\mathrm{Ker}(v))\oplus\mathrm{Im}(v)$.

To see the sufficiency of the first point, write down the matrix for $u+\lambda v$ where the initial base of $V$ is adapted to $S\oplus \mathrm{Ker}(u)=V$, and the final base is the images under $u$ of the basis vectors chosen in $S$ and the images under $v$ of the basis vectors in $\mathrm{Ker}(u)$, and let $\lambda\rightarrow 0$: the determinant of this matrix is a polynomial in $\lambda$ of the form $\lambda^d(1+\cdots)$ where $d=\dim~\mathrm{Ker}(u)$.

Going back to the initial problem, we get $3$ sufficient conditions for the span of $u$ and $v$ to be completely singular, and $2$ necessary conditions.

Solution 2:

A sufficient condition is $P$ and $Q$ share at least one eigenvector for 0 eigenvalue:

$P\mathbf{v}= 0 \mathbf{v}$, $Q\mathbf{v}= 0 \mathbf{v}$

so for any $\lambda,\mu\in\mathbb{R}$, we have $(\lambda P+\mu Q) \mathbf{v} = 0 \mathbf{v}$, which means 0 is still an eigenvalue of $\lambda P+\mu Q$, i.e. $\det(\lambda P+\mu Q)=0$.

I doubt it's also a necessary condition.